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UVA 113 Power of Cryptography

2014年09月05日 ⁄ 综合 ⁄ 共 1108字 ⁄ 字号 评论关闭

 Power of Cryptography 


Background

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics
once considered to be of only theoretical interest.

This problem involves the efficient computation of integer roots of numbers.

The Problem

Given an integer tex2html_wrap_inline32 and an integer tex2html_wrap_inline34 you
are to write a program that determines tex2html_wrap_inline36 , the positive tex2html_wrap_inline38 root
of p. In this problem, given such integers n and pp will always be of the form tex2html_wrap_inline48 for an integer k (this
integer is what your program must find).

The Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs tex2html_wrap_inline56 ,tex2html_wrap_inline58 and
there exists an integer ktex2html_wrap_inline62 such that tex2html_wrap_inline64 .

The Output

For each integer pair n and p the value tex2html_wrap_inline36 should be printed, i.e., the number k such that tex2html_wrap_inline64 .

Sample Input

2
16
3
27
7
4357186184021382204544

Sample Output

4
3
1234

想的太复杂了,刚开始还想用字符串来着,但是在网上找相关解法知道用double可解决问题,double大小可达到2的1024次方。
但是强制类型转换时要注意精度问题,通常方法是+0.5之后再转换类型。
代码如下:
#include <stdio.h>
#include <math.h>
int main(void)
{
    double n,p;
    while(scanf("%lf%lf",&n,&p)!=EOF)
     printf("%.0f\n",pow(p,1.0/n));
    return 0;
}

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